Problem: Steph makes $90\%$ of the free throws she attempts. She is going to shoot $3$ free throws. Assume that the results of free throws are independent from each other. Let $X$ represent the number of free throws she makes. Find the probability that Steph makes exactly $1$ of the $3$ free throws. You may round your answer to the nearest hundredth. $P(X=1)=$
Without a fancy calculator Making $1$ free throw in $3$ attempts means Steph needs to make $1$ free throw and miss $2$ free throws. For each free throw, we know $P({\text{make}})={90\%}$ and $P({\text{miss}})={10\%}$. Since we are assuming independence, let's multiply probabilities to find the probability of making $1$ attempt followed by missing $2$ attempts — ${\text{S}}$ is a free throw she makes and ${\text{F}}$ is a free throw she misses. $P({\text{S}}{\text{FF}})=({0.9})({0.1})^2=0.009$ This isn't our final answer though, because there are other ways to get $1$ make in $3$ attempts (for example, FSF). How many different ways are there? We can use the combination formula to find how many ways there are to get $1$ make in $3$ attempts: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _3\text{C}_1&=\dfrac{3!}{(3-1)!\cdot1!} \\\\ &=\dfrac{3 \cdot \cancel{2 \cdot 1}}{(\cancel{2 \cdot 1}) \cdot 1 } \\\\ &=3 \end{aligned}$ There are $3$ ways she can make $1$ shot in $3$ attempts. Do they all have the same probability? Each of the $3$ ways has the same probability that we already found: $\begin{aligned} P({\text{S}}{\text{FF}})&=({0.9})({0.1})^2=0.009 \\\\ P({\text{F}}{\text{S}}{\text{F}})&=({0.9})({0.1})^2=0.009 \\\\ P({\text{FF}}{\text{S}})&=({0.9})({0.1})^2=0.009 \end{aligned}$ So we can multiply this probability by $3$ since that is how many ways there are to make $1$ shot in $3$ attempts. $\begin{aligned} P(X=1)&=3(0.9)(0.1)^2 \\\\ &=3 \cdot 0.009 \\\\ &=0.027 \\\\ &\approx0.03 \end{aligned}$ $P(X=1)=0.027\approx0.03$